Question

(a) A small object is placed 150 mm away from a diverging lens of focal length 100 mm.
(i) Copy the figure below and draw rays to show how an image is formed by the lens.
Figure
(ii) Calculate the distance of the image from the lens by using the lens formula.
(b) The diverging lens in part (a) is replaced by a converging lens also of focal length 100 mm. The object remains in the same position and an image is formed by the converging lens. Compare two properties of this image with those of the image formed by the diverging lens in part (a).

Answer 1

(a)(i) The image formed by the lens is virtual, erect and diminished in size.


(a)(ii) Object distance (u) = -150 mm = -15 cm (sign convention)
Focal length (f) = -100 mm = -10 cm (sign convention)
Image distance (v) = ?

Applying the lens formula, we have:

$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{v}-\frac{1}{-15}=\frac{1}{-10} \\ \Rightarrow \frac{1}{v}+\frac{1}{15}=-\frac{1}{10} \\ \Rightarrow \frac{1}{v}=-\frac{1}{10}-\frac{1}{15} \\ \Rightarrow \frac{1}{v}=\frac{-3-2}{30} \\ \Rightarrow \frac{1}{v}=\frac{-5}{30} \\ \Rightarrow v=-6\mathrm{cm}. \end{gathered}$$


(b)

S.No	Image formed by a diverging lens	Image formed by a converging lens
1	Image formed is virtual and erect	Image formed is real and inverted
2	Image is smaller in size than the object	Image is larger in size than the object